Continuum set into aleph zero many subsets How to prove that if a set with cardinality of continuum  is divided into aleph zero many subsets, then at least one of them has cardinality of continuum? (Of course without assuming the Continuum hypothesis)
Edit: Not precise formulation.
 A: EDIT: Now that you have clarified your question, I can answer it. The answer to this, perhaps surprisingly, is yes, regardless of whether CH holds! This follows from Koenig's Theorem, which implies that $cf(2^{\aleph_0})>\omega$: if $\mathbb{R}=\bigcup_{i\in\mathbb{N}}A_i$, let $\alpha_i=\vert A_1\cup A_2\cup . . . \cup A_i\vert$; then $\sup\alpha_i=2^{\aleph_0}$, but if each $A_i$ has size $<2^{\aleph_0}$, each $\alpha_i$ is $<2^{\aleph_0}$ (exercise), so the sequence of $\alpha_i$s is a length-$\omega$ sequence cofinal in $2^{\aleph_0}$ - a contradiction.
Note that choice is necessary here: in fact, without choice it is consistent that $\mathbb{R}$ is the union of two sets of strictly smaller cardinality!

EDIT 2: Here's a direct proof, without invoking Koeneig's theorem (although the argument below basically is the proof of Koenig's theorem):
Remember that the set $\mathbb{R}^\mathbb{N}$ of functions from $\mathbb{N}$ to $\mathbb{R}$ also has size continuum; we'll show that if we partition this set into countably many pieces, one must have size continuum.
Suppose otherwise. Let $A_i\subset \mathbb{R}^\mathbb{N}$ have size $<2^{\aleph_0}$, for $i\in\mathbb{N}$. Then for each $i$, there is some real $r_i$ such that whenever $s\in A_i$, we have $s(i)\not=r_i$. (Why not?) But now consider the function $$f(i)=r_i.$$ This $f$ is an element of $\mathbb{R}^\mathbb{N}$, but not in any $A_i$.
A: Here is a fairly direct argument (this is really the same as Noah's answer; I'm just implementing the proof of König's theorem in this particular case).  Suppose $X=\bigcup_{n\in\mathbb{N}} X_n$ has cardinality $\mathfrak{c}$ but each $X_n$ has cardinality less than $\mathfrak{c}$.  Since $\mathfrak{c}^{\aleph_0}=\mathfrak{c}$, there exists a bijection $f:X\to X^\mathbb{N}$.  For each $n$, let $s_n\in X$ be some element which is not the $n$th coordinate of $f(x)$ for any $x\in X_n$ (such an $s_n$ exists since $|X_n|<|X|$).  Now consider the sequence $s=(s_n)\in X^{\mathbb{N}}$.  Since $f$ is surjective, $s=f(x)$ for some $x\in X$, and $x\in X_n$ for some $n$.  But by construction, the $n$th coordinate of $s$ is then different from the $n$th coordinate of $f(x)$.  This is a contradiction.
